 Hello and welcome to the session. Let's discuss the following question. It says verified Lagrange's mean value theorem for the function fx is equal to under the root x squared minus 4 in the interval 2 to 4. So let's now move on to the solution fx is under the root x squared minus 4 and it is defined on the closed interval 2 to 4 Now for the verification of the Lagrange's mean value theorem we need to verify that the given function should be continuous on the closed interval and it should be differentiable on the open interval and there exists a point c belonging to that interval so that f dash c is equal to fb minus f a upon b minus a where is the initial point of the interval and b is the end point of the interval now fx is under the root x square minus 4 that means fx has a unique and definite value for each value of the interval 2 to 4 that is fx attains a unique and definite value for each x belonging to the interval 2 to 4 therefore fx is continuous on the closed interval 2 to 4 Now we have to find f dash x that is we have to check that the given function is Differentiable on the open interval 2 to 4 or not now f dash x is given by 1 by 2 into x square minus 4 to the power 1 by 2 minus 1 into the derivative of x square minus 4 which is 2x so this is equal to 2x upon 2 into x square minus 4 to the power 1 by 2 so this is equal to x upon under the root x square minus 4 and we see that f dash x exists for all values of x belonging to the interval 2 to 4 therefore f dash x exists and is finite for each x Belonging to the open interval 2 to 4 so this implies fx is Differentiable on the open interval 2 to 4 so thus the conditions of Lagrange's Mean value theorem are satisfied therefore there exist See belonging to the open interval 2 to 4 Such that dash c is equal to fb. Here b is 4 Minus f a a is 2 upon b minus a that is 4 minus 2 now Fb is That is f of 4 is under the root 4 square minus 4 Minus f2 so we'll put x is equal to 2 here. So we have 2 square minus 4 upon 2 so this is equal to 16 minus 4 minus 4 minus 4 upon 2 So this is equal to 16 minus 4 is 12. So this is under the root 12 by 2 4 minus 4 is 0. So we have f dash c is equal to root 3 by 2 Do 12 can be written as 2 root 3 2 gets cancelled with 2 and we have f dash c is c upon under the root c Square minus 4 so we have c upon under the root c square minus 4 is equal to under the root 3 So this implies c square is equal to 3 into C square minus 4 taking square on both sides we have Now c square is equal to 3 c square minus 4 so this implies C square minus 3 c square is equal to minus 4. So this implies minus 2 c square is equal to minus 4. So this implies C square is equal to 2 So this implies C is equal to now here 3 into 4 is 12. So here we need to have 12 So it is 3 c square minus 12 and Here also it is minus 12 minus 12 and c square is equal to 6 So this implies c is equal to plus minus under the root 6. Now we see that C is equal to under the root 6 that is positive square root belongs to the interval 2 to 4 Thus there exists c belonging to the open interval 2 to 4 such that f dash c is equal to Fb minus Fa upon b minus a thus Hence Lagrange's mean value theorem is very fight So this completes the question and the session. Bye for now. Take care. Have a good day